Question

From the following molar conductivities at infinite dilutions

^°m for Ba(OH)2 = 457.6 ohm-1cm2mol-1

^°m for Bacl2= 240.6 ohm-1cm2mol-1

^°m for NH4cl = 129.8 ohm-1cm2mol-1

Calculate the ^°m of NH4cl

Answer 1

$\lambda$°_m of Ba(OH)₂ is 457.6 ohm^-1*cm²*mol^-1;

$\lambda$°_m of BaCl₂ is 240.6 ohm^-1*cm²*mol^-1;

$\lambda$°_m of NH₄Cl is 129.8 ohm^-1*cm²*mol^-1;

The $\lambda$°_m of NH₄Cl is 129.8 cm²*ohm^-1*mol^-1;

Probably should calculated the $\lambda$°_m of NH₄OH.

Then, NH₄OH = NH₄⁺ + OH^-;

So, $\lambda$°_m(NH₄OH) = $\lambda$°_m(NH₄⁺) + $\lambda$°_m(OH^-);

Or $\lambda$°_m(NH₄OH) = $\lambda$°_m(NH₄Cl) - (1/2 * $\lambda$°_m(BaCl₂)) + (1/2 * $\lambda$°_m(Ba(OH)₂)) = (($\lambda$°_m(NH₄⁺) + $\lambda$°_m(Cl^-)) - (1/2 * ( $\lambda$°_m(Ba²⁺) + 2 * $\lambda$°_m(Cl^-))) + (1/2 * ( $\lambda$°_m(Ba²⁺) + 2 * $\lambda$°_m(OH^-))) = $\lambda$°_m(NH₄⁺) + $\lambda$°_m(Cl^-) - 1/2 * $\lambda$°_m(Ba²⁺) - $\lambda$°_m(Cl^-) + 1/2 * $\lambda$°_m(Ba²⁺) + $\lambda$°_m(OH^-) = $\lambda$°_m(NH₄⁺) + $\lambda$°_m(OH^-).

So:

$\lambda$°_m(NH₄OH) = $\lambda$°_m(NH₄Cl) - (1/2 * $\lambda$°_m(BaCl₂)) + (1/2 * $\lambda$°_m(Ba(OH)₂) = 129.8 - (1/2 * 240.6) + (1/2 * 457.6) = 129.8 - 120.3 + 228.8 = 238.3 cm²*ohm^-1*mol^-1.

Answer: 238.3 cm²*ohm^-1*mol^-1.